Differentiate the following integral 
Let the following integral
$$I(y)=2\int_{\sqrt {\frac 1 y -1}}^{\infty} \frac 1 {\pi} \frac 1 {1+t^{2}} dt.$$
Compute $\frac{dI}{dy}$

I'm trying to use Leibniz's formula but the limit $\infty$ confuses me a little, I don't know if this procedure is the correct one or maybe there is a better way to calculate, any suggestion or help would be much appreciated.
 A: This is actually a neat problem, because you can perform the procedure of differentiating the integral in two equivalent ways and check the result.
The first is to actually evaluate the integral.  Recall that $d\tan^{-1}(t)/dt = 1/(1+t^2)$.  Thus $$I(y) = 1-\frac2\pi\tan^{-1}\left( \sqrt{\frac1y-1} \right),$$ since $\tan^{-1}(\infty)= \pi/2$. Then, we may explicitly evaluate $dI/dy$ by repeated use of the chain rule, $$\frac{dI}{dy} = -\frac2\pi\frac{1}{1+\left( \frac1y-1 \right)}\frac12\frac{1}{\sqrt{\frac1y-1}}\frac{-1}{y^2} = \frac{1}{\pi\, y}\frac{1}{\sqrt{\frac1y-1}}.$$
The second method is to apply Leibniz' Rule, which helps us when differentiating with respect to the limits and/or integrand inside an integral; see, e.g., https://en.wikipedia.org/wiki/Leibniz_integral_rule.  In the case at hand, only the lower bound of the integral depends on $y$, so we only need to evaluate $$\frac{dI}{dy} = \frac2\pi\left.\frac{1}{1+t^2}\right|_{t=\sqrt{\frac1y-1}}\frac{d}{dy}\sqrt{\frac1y-1} = \frac{1}{\pi \, y}\frac{1}{\sqrt{\frac1y-1}}.$$
